6

Question from my linear algebra homework I'm struggling with:

Let $D = \begin{bmatrix} -2 & 5 & 4 \\-1 & 0 & 0 \\0 & 4 & 3 \end{bmatrix}$

We are asked:

  1. Find $D^5+3D^2-D+I$

  2. Find $D^{2013}$

  3. Write $D^{-1}$ as a polynomial of $D$

I solved questions 1) and 3) but can't solve 2)...

Oria Gruber
  • 13,035

5 Answers5

11

Here is how I would attack such a question. You need to find a polynomial equation satisfied by$~D$ first. You could use the characteristic polynomial for that (by the Cayley-Hamilton theorem), but knowing that such a polynomial of degree at most$~3$ exists, you can also just try to find a relation between some powers of$~D$. The sparse second line suggests right-multiplying powers of $D$ to $(0~1~0)$, giving the sequence $(0~1~0),(-1~0~0),(2~{-}5~{-}4),(3~{-}6~{-}4)$. The first three are clearly linearly independent, and the fourth one gives the relation $(0~1~0)-(-1~0~0)-(2~{-}5~{-}4)+(3~{-}6~{-}4)=(0~0~0)$, so your polynomial equation should be $I-D^1-D^2+D^3=0$, which you can check to be true. Now $X^3-X^2-X+1=(X-1)(X^2-1)=(X-1)^2(X+1)$. Indeed your matrix is not diagonalisable because of the double root.

To compute $X^{2013}$ you can take its remainder$~R$ after division by $P=(X-1)(X^2+1)$, so that $X^{2013}=PQ+R$ for some (quotient)$~Q$. For finding the remainder after division by a polynomial with such easy (complex) roots as $P$ has, the standard trick to avoid doing a (very) long division of polynomials is write the remainder as a polynomial of degree${}<\deg P=3$ with unknown coefficients: $R=aX^2+bX+c$, and evaluate the equation $X^{2013}=PQ+R$ at the roots of$~P$; since these substitutions annihilate the term $PQ$ regardless of$~Q$, they give you linear equations in $a,b,c$. The problem here is that you have only two roots $1,-1$ to substitute, although $1$ is a double root of the minimal polynomial. There is another trick to solve this shortage of equations: since $1$ is also a root of the derivative $P'$ of$~P$, you can take the derivative of the equation (being an identity of polynomials in$~X$, this gives an equation that must still hold) giving $2013X^{2012}=P'Q+PQ'+R'$. Now substitute $X=1$ into that to get a third equation that will let you solve $a,b,c$. The answer is $aD^2+bD+cI$.

Marc van Leeuwen
  • 119,547
  • Thank you very much. That is indeed the way the question should be solved. I didn't think of deriving. Thanks. top answer. – Oria Gruber Nov 24 '13 at 11:25
5

Although $D$ is not diagonalizable, it has the Jordan decomposition (courtesy of Wolfram Alpha): $$D = SJS^{-1}$$ where

$$S = \begin{bmatrix} -1 & 1 & \frac12 \\-1 & -1 & \frac12 \\1 & 2 & 0 \end{bmatrix}$$

and

$$J = \begin{bmatrix} -1 & 0 & 0 \\0 & 1 & 1 \\0 & 0 & 1 \end{bmatrix}$$

There is a simple closed-form expression for $J^n$, which you should be able to find and prove yourself.

TonyK
  • 68,059
  • I think the point is to use polynomial division and cayley-hamilton theorem, we didnt study jordan form yet so i dont think thats the way – Oria Gruber Nov 24 '13 at 10:52
  • 1
    Be careful about what you study, there aren't always epiphany answers. Jordan normal form is an amazing tool. – DanielV Sep 05 '15 at 18:46
5

The characteristic polynomial of $D$ is given by

$$\chi_D(\lambda) = \det(\lambda I_3 - D) = \lambda^3 - \lambda^2 - \lambda + 1 = (\lambda - 1)^2 (\lambda+1)$$

By Cayley Hamilton theorem, we have

$$\chi_D(D) = D^3 - D^2 - D + I_3 = (D - I_3)^2(D + I_3) = (D^2 - I_3)(D - I_3) = 0$$

Since $(\lambda + 1)\chi_D(\lambda) = (\lambda^2 - 1)^2$, we have $(D^2 - I_3)^2 = 0$ and hence $$\begin{align} D^{2013} = & (I_3 + (D^2 - I_3))^{1006} D\\ = & (I_3 + 1006 (D^2 - I_3)) D\quad\color{blue}{\longleftarrow \text{throwing away terms}\propto (D^2 - I_3)^2}\\ = & D + 1006 (D^2 - I_3)((D - I_3) + I_3)\\ = & D + 1006 (D^2 - I_3)\quad\color{blue}{\longleftarrow \text{throwing away terms}\propto (D^2 - I_3)(D-I_3)}\\ = & \left[\begin{array}{rrr} -2014 & 6041 & 4028 \\ 2011 & -6036 & -4024 \\ -4024 & 12076 & 8051 \end{array}\right] \end{align} $$

achille hui
  • 125,323
  • Looks all right, but you might bring attention to the fact that in the before-last equality you applied the identity $D^3-D=D^2-I$ that follows from the original polynomial equation. Without that, it looks like you did the algebra wrong. – Marc van Leeuwen Nov 24 '13 at 11:46
  • @MarcvanLeeuwen thanks for point out that. I've fixed that together with other minus sign errors in the original derivation. – achille hui Nov 24 '13 at 12:09
2

Bull headed approach. 2013 written in binary form is 11111011101.

Therefore, repeatedly square D 10 times which gives you $D$, $D^2$, $D^4$, $D^8$, $D^{16}$, $D^{32}$, $D^{64}$, $D^{128}$, $D^{256}$, $D^{512}$ and $D^{1024}$.

Then multiply $D^{1024} \times D^{512} \times D^{256} \times D^{128} \times D^{64} \times D^{16} \times D^{8} \times D^4$.

So, you will have to perform 17 multiplications, which is doable.

TenaliRaman
  • 3,916
  • 1
    If you've ever computed $n^{1024}$ for an integer $n\neq{1,0,-1}$, there is no reason to be optimistic about computing $D^{1024}$. It happens that this is a very special matrix, and the coefficients of its powers grow only linearly with the exponent. But I don't think there is reason to conclude "doable" without having done it. – Marc van Leeuwen Nov 24 '13 at 11:30
  • Sure, give me sometime and I will actually do it and write it up. Do you want me to? My point that it was doable precisely because it is a special matrix. – TenaliRaman Nov 24 '13 at 11:32
0

Ok, I found the way to answer this, but I've still run into some trouble.

Basically the Idea is this:

you take the polynomial $t^{2013}$ and you divide it by the characteristic polynomial which is $-t^3+t^2+t-1$ and you get:

$t^{2013} = (-t^3+t^2+t-1)*q(t)+r(t)$ where $deg(r(t)) \leq deg(-t^3+t^2+t-1)-1 = 2$

$t=1$ and $t=-1$ are roots of $-t^3+t^2+t-1$ so if $r(x)=ax^2+bx+c$:

$1^{2013} = a+b+c$ and $(-1)^{2013} = a-b+c$

If we find a b c then we solved the question. But how do we find a b c?

Oria Gruber
  • 13,035